Introduction to Open Data Science - Course Project

About the project

Write a short description about the course and add a link to your GitHub repository here. This is an R Markdown (.Rmd) file so you should use R Markdown syntax.

Course description

The name of this course refers to THREE BIG THEMES: 1) Open Data, 2) Open Science, and 3) Data Science. These themes are summarized briefly as follows:

  1. Open Data

There are more and more open data sets available. Utilizing and sharing data is an essential skill for researchers in all fields. During this course we use open data sets from different sources and learn to prepare them for different analyses. You will explore, analyse and interpret data from real world applications.

  1. Open Science

Science thrives to be open. Repeating or reproducing the results is a common aim in any branch of science, but it is not always easy or simple. Sharing data is not enough for reproducibility. What is also needed, is using openly available software tools and methods as well as sharing your code and results. You will learn these skills during this course, using state-of-the-art tools.

  1. Data Science

Data Science is the name for the data driven world of Statistics. Nowadays, finding or collecting data is not a problem. Instead, the challenges are in extracting knowledge and discovering the patterns behind the data. It requires skills of coding, programming, and modelling, as well as visualizing and analysing. You will face all these topics on this course.

Contents of the course:

1 Tools and methods for open and reproducible research
R, RStudio, R Markdown, GitHub
2 Regression and model validation
3 Logistic regression
4 Clustering and classification
Discriminant analysis
K-means clustering
5 Dimensionality reduction techniques
Principal component analysis
Multiple correspondence analysis
6 Analysis of longitudinal data
Graphical displays and summary measures
Linear mixed effects models

After completing this course we will understand the principles and advantages of using open research tools with open data and understand the possibilities of reproducible research. We will know how to use R, RStudio, R Markdown, and GitHub for these tasks and also know how to learn more of these open software tools. We will also know how to apply certain statistical methods of data science, that is, data-driven statistics.

The link to my IODS GitHub repository is: https://github.com/tammisi/IODS-project.git

My thoughts about the course

I’m looking forward to the course and to learning to use R and Github better at my daily work. I also expect to get a better understanding of performing basic statistical tests. I have been using bioinformatic tools, mostly R, in my PhD project, but I want to learn to be more efficient and organized in my work. I am also just starting with regression analysis, so this course comes at a very good time. So alltogether I have big expectations for the course!

Reflection on learning experience with excercise set 1

I am familiar with the basics of R, so there wasn’t anything very new for me in exercise set 1. But it is always nice to repeat to refresh and enhance your memory. It was especially good to get some clarity on how to for example filter and modify the data, since there isn’t always a good logic in the way I use all the tools and functions. So one goal for me in this course is to learn to be more efficient and systematic in my R scripts. I haven’t been working with dates much in my previous work with R, so there were also some useful tips about that topic that can come in handy in the future! I’m also looking forward to the exercises on statistical analysis, because I am a novice in that area, but I am just starting to need them in my work.

Chapter2, Analysis part

Describe the work you have done this week and summarize your learning.

date()
## [1] "Sun Dec 10 17:18:41 2023"
# libraries needed in this exercise
library(tidyverse)
library(readr)
library(ggplot2)
library(GGally)

Exercise 1

Read the students2014 data into R. Explore the structure and the dimensions of the data and describe the dataset briefly, assuming the reader has no previous knowledge of it.

# read in the 
learning2014 <- read_csv("./data/learning2014.csv")
## 
## ── Column specification ────────────────────────────────────────────────────────
## cols(
##   gender = col_character(),
##   age = col_double(),
##   attitude = col_double(),
##   deep = col_double(),
##   stra = col_double(),
##   surf = col_double(),
##   points = col_double()
## )
dim(learning2014)
## [1] 166   7
str(learning2014)
## spc_tbl_ [166 × 7] (S3: spec_tbl_df/tbl_df/tbl/data.frame)
##  $ gender  : chr [1:166] "F" "M" "F" "M" ...
##  $ age     : num [1:166] 53 55 49 53 49 38 50 37 37 42 ...
##  $ attitude: num [1:166] 3.7 3.1 2.5 3.5 3.7 3.8 3.5 2.9 3.8 2.1 ...
##  $ deep    : num [1:166] 3.58 2.92 3.5 3.5 3.67 ...
##  $ stra    : num [1:166] 3.38 2.75 3.62 3.12 3.62 ...
##  $ surf    : num [1:166] 2.58 3.17 2.25 2.25 2.83 ...
##  $ points  : num [1:166] 25 12 24 10 22 21 21 31 24 26 ...
##  - attr(*, "spec")=
##   .. cols(
##   ..   gender = col_character(),
##   ..   age = col_double(),
##   ..   attitude = col_double(),
##   ..   deep = col_double(),
##   ..   stra = col_double(),
##   ..   surf = col_double(),
##   ..   points = col_double()
##   .. )

Answer:
The dataset has 166 rows and 7 columns. The data consists of results from a survey on student’s attitudes towards statistics and their exam points. In the data we have variables like gender and age, as well as attitude, which is student’s global attitude toward statistics, and also student’s answers to questions that have bee grouped by their relatedness to deep (deep), surface (surf) and strategic (stra) learning and that have been scaled back to the original 1-5 scale by taking the mean.

Exercise 2

Show a graphical overview of the data and show summaries of the variables in the data. Describe and interpret the outputs, commenting on the distributions of the variables and the relationships between them.

# get summary of variables
summary(learning2014)
##     gender               age           attitude          deep      
##  Length:166         Min.   :17.00   Min.   :1.400   Min.   :1.583  
##  Class :character   1st Qu.:21.00   1st Qu.:2.600   1st Qu.:3.333  
##  Mode  :character   Median :22.00   Median :3.200   Median :3.667  
##                     Mean   :25.51   Mean   :3.143   Mean   :3.680  
##                     3rd Qu.:27.00   3rd Qu.:3.700   3rd Qu.:4.083  
##                     Max.   :55.00   Max.   :5.000   Max.   :4.917  
##       stra            surf           points     
##  Min.   :1.250   Min.   :1.583   Min.   : 7.00  
##  1st Qu.:2.625   1st Qu.:2.417   1st Qu.:19.00  
##  Median :3.188   Median :2.833   Median :23.00  
##  Mean   :3.121   Mean   :2.787   Mean   :22.72  
##  3rd Qu.:3.625   3rd Qu.:3.167   3rd Qu.:27.75  
##  Max.   :5.000   Max.   :4.333   Max.   :33.00
# draw a scatter plot matrix of the variables in learning2014.
# [-1] excludes the first column (gender)
pairs(learning2014[-1])

# create a more advanced plot matrix with ggpairs()
ggpairs(learning2014, mapping = aes(), lower = list(combo = wrap("facethist", bins = 20)))

Answer:
In the summary we can see that age of the participants ranges between 17 and 55, mean being 26. Attitudes on 1-5 scale range from 1.4 to 5.0, mean is 3.1. Exam points range from 7.0 to 33.0 and the mean is 22.7.

From the plot matrix we can see that attitude correlates strongly with points (cor=0.437) and also answers related to surface and deep learning correlate with each other.

Exercise 3

Exercise 4

Using a summary of your fitted model, explain the relationship between the chosen explanatory variables and the target variable (interpret the model parameters). Explain and interpret the multiple R-squared of the model

Answer:
After removing the non-significant variables, attitude is still statistically significant (p-value now 4.12e-09). The predicted change in exam points per one unit change in attitude is 3.5255. Standard error is 0.57 so pretty low. T-value also indicates that the estimated coefficient is 6.2 standard errors away from 0. The model now explains 18.6% (adjusted R-squared = 0.1856) of the variation in exam points. So the results didn’t change much when removing the non-significant variables.

Exercise 5

Produce the following diagnostic plots: Residuals vs Fitted values, Normal QQ-plot and Residuals vs Leverage. Explain the assumptions of the model and interpret the validity of those assumptions based on the diagnostic plots

par(mfrow=c(2,2))
plot(learning.lm, which=c(1,2,5))

Answer:
Residuals vs. fitted plot show that the points are quite nicely evenly distributed and don’t show any clear pattern, show the model seem to fit the data well. No modifications such as quadratic terms, are needed. Also the QQ plot showing the residuals against the same from a normal distribution look like a pretty even diagonal line, except some small deviations in the lower and higher ends, so the data seems to be normally distributed. The residuals vs leverage plot doesn’t show any strong outliers that would have a significant effect on the results. So according to the diagnostic plots, the model seems to perform well on the data.

Chapter3, Analysis part

Exercise 1

Read the joined student alcohol consumption data into R

alc <- read.csv("./data/alc.csv")
dim(alc)
## [1] 370  36
head(alc)
##   X school sex age address famsize Pstatus Medu Fedu     Mjob     Fjob
## 1 1     GP   F  18       U     GT3       A    4    4  at_home  teacher
## 2 2     GP   F  17       U     GT3       T    1    1  at_home    other
## 3 3     GP   F  15       U     LE3       T    1    1  at_home    other
## 4 4     GP   F  15       U     GT3       T    4    2   health services
## 5 5     GP   F  16       U     GT3       T    3    3    other    other
## 6 6     GP   M  16       U     LE3       T    4    3 services    other
##       reason guardian traveltime studytime schoolsup famsup activities nursery
## 1     course   mother          2         2       yes     no         no     yes
## 2     course   father          1         2        no    yes         no      no
## 3      other   mother          1         2       yes     no         no     yes
## 4       home   mother          1         3        no    yes        yes     yes
## 5       home   father          1         2        no    yes         no     yes
## 6 reputation   mother          1         2        no    yes        yes     yes
##   higher internet romantic famrel freetime goout Dalc Walc health failures paid
## 1    yes       no       no      4        3     4    1    1      3        0   no
## 2    yes      yes       no      5        3     3    1    1      3        0   no
## 3    yes      yes       no      4        3     2    2    3      3        2  yes
## 4    yes      yes      yes      3        2     2    1    1      5        0  yes
## 5    yes       no       no      4        3     2    1    2      5        0  yes
## 6    yes      yes       no      5        4     2    1    2      5        0  yes
##   absences G1 G2 G3 alc_use high_use
## 1        5  2  8  8     1.0    FALSE
## 2        3  7  8  8     1.0    FALSE
## 3        8 10 10 11     2.5     TRUE
## 4        1 14 14 14     1.0    FALSE
## 5        2  8 12 12     1.5    FALSE
## 6        8 14 14 14     1.5    FALSE

Exercise 2

Exercise 3

The purpose of your analysis is to study the relationships between high/low alcohol consumption and some of the other variables in the data. To do this, choose 4 interesting variables in the data and for each of them, present your personal hypothesis about their relationships with alcohol consumption.

I choose variables sex, school, failures and famrel.
H0 hypothesis is that the variables have no connection to alcohol consumption among students.
My H1 hypotheses are:
- sex is connected to high alcohol use (for sex=M use is higher).
- there might be differences between the schools
- higher number of past class failures and poorer quality of family relationships is connected to higher alcohol consumption.

Exercise 4

Numerically and graphically explore the distributions of your chosen variables and their relationships with alcohol consumption (use for example cross-tabulations, bar plots and box plots). Comment on your findings and compare the results of your exploration to your previously stated hypotheses.

# bar plots of all variables
mydata <- select(alc, c(high_use, school, sex, famrel, failures))
gather(mydata) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") + geom_bar()

# barplots of non-numerical variables
g1 <- ggplot(alc, aes(x=high_use)) + geom_bar() + facet_wrap('sex') + ggtitle('high_use vs sex') + theme(plot.title = element_text(hjust = 0.5))

g2 <- ggplot(alc, aes(x=high_use)) + geom_bar() + facet_wrap('school') + ggtitle('high_use vs school') + theme(plot.title = element_text(hjust = 0.5))

# boxplots of numerical variables
g3 <- ggplot(alc, aes(x=high_use, y=famrel)) + geom_boxplot() + ggtitle('high_use vs famrel') + theme(plot.title = element_text(hjust = 0.5))

ggarrange(g1,g2,g3)

#g4 <- ggplot(alc, aes(x=high_use, y=failures)) + geom_boxplot() + ggtitle('high_use vs failures') + theme(plot.title = element_text(hjust = 0.5)) 
#g4
# for some reason this doesn't work for failures, so I use boxplot in this case instead:
boxplot(alc$high_use, alc$failures, xlab='high use', ylab='failures', main='high_use vs. failures', names=levels(factor(alc$high_use)))

From the first barplot we can see that most students have zero and only very few have 1-3 past failures. Most of the students also have good family relations. High alcohol use is more uncommon than low alcohol use. There are a lot more students from school ‘GP’ than ‘MS’, but the ratio between female and male students is pretty even.

The barplot of high_use vs. sex indicates that larger proportion of male students are high users of alcohol, so this indicates that the hypotheses could be true.

The barplot of high_use vs. school indicates that in both schools the proportion of high users is lower than not high users. Even though the difference is bigger in GP, the difference might not be statistically significant.

The barplot of high_use vs. famrel indicates that the mean quality of family relations is lower among high users. The hypothesis could therefore be true.

The barplot of high_use vs. failures indicates that both high and low users have very few past failures. Althoug there are some individuals with higer amounts of failures among high users, I don’t think this difference is going to be statistically significant, so the hypothesis migh not hold.

Exercise 5

Use logistic regression to statistically explore the relationship between your chosen variables and the binary high/low alcohol consumption variable as the target variable

# fit model
alc.glm <- glm(high_use ~ sex + school + failures + famrel, data = alc, family = "binomial")
summary(alc.glm)
## 
## Call:
## glm(formula = high_use ~ sex + school + failures + famrel, family = "binomial", 
##     data = alc)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.7345  -0.8246  -0.6257   1.1687   1.9935  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  -0.2982     0.5170  -0.577 0.564098    
## sexM          0.9357     0.2415   3.874 0.000107 ***
## schoolMS      0.3071     0.3725   0.824 0.409704    
## failures      0.6162     0.2018   3.054 0.002260 ** 
## famrel       -0.3083     0.1276  -2.415 0.015718 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 452.04  on 369  degrees of freedom
## Residual deviance: 418.96  on 365  degrees of freedom
## AIC: 428.96
## 
## Number of Fisher Scoring iterations: 4
confint(alc.glm)
## Waiting for profiling to be done...
##                  2.5 %      97.5 %
## (Intercept) -1.3226184  0.71249519
## sexM         0.4671197  1.41565633
## schoolMS    -0.4438613  1.02622019
## failures     0.2278800  1.02645318
## famrel      -0.5609504 -0.05862812
coef(alc.glm) %>% exp
## (Intercept)        sexM    schoolMS    failures      famrel 
##   0.7421802   2.5490315   1.3594796   1.8518843   0.7346998

Present and interpret a summary of the fitted model. Present and interpret the coefficients of the model as odds ratios and provide confidence intervals for them. Interpret the results and compare them to your previously stated hypothesis.

Seems like School is not a signifant variable explaining alcohol consumption (p-value 0.409704). However,sex is a highly significant variable (p-value 0.000107). The odds of student being a high user is 2.6 times higher (exp(0.94) = 2.55) if he is a male compared with a female, if all other variables are constant.Failures is also significant (p-value 0.002260), one unit increase in number of failures increases log odds of high use by 0.62 (95% CI 0.23, 1.03) units. Famrel is also somewhat significant (p-value 0.015718), every one unit increase in the quality of family relations decreases log odds of high use by -0.31 (95% CI -0.56, -0.06) units. so the hypotheses for sex, failures and famrel were correct.

Exercise 6

Using the variables which, according to your logistic regression model, had a statistical relationship with high/low alcohol consumption, explore the predictive power of you model. Provide a 2x2 cross tabulation of predictions versus the actual values and optionally display a graphic visualizing both the actual values and the predictions. Compute the total proportion of inaccurately classified individuals (= the training error) and comment on all the results. Compare the performance of the model with performance achieved by some simple guessing strategy

# retrain model with statistically significant variables
alc.glm <- glm(high_use ~ sex + failures + famrel, data=alc, family='binomial')
summary(alc.glm)
## 
## Call:
## glm(formula = high_use ~ sex + failures + famrel, family = "binomial", 
##     data = alc)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -1.7465  -0.8333  -0.6359   1.1559   1.9814  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  -0.2321     0.5109  -0.454 0.649564    
## sexM          0.9322     0.2413   3.863 0.000112 ***
## failures      0.6144     0.2021   3.040 0.002366 ** 
## famrel       -0.3159     0.1276  -2.476 0.013273 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 452.04  on 369  degrees of freedom
## Residual deviance: 419.63  on 366  degrees of freedom
## AIC: 427.63
## 
## Number of Fisher Scoring iterations: 4
# make predictions
probabilities <- predict(alc.glm, type = "response")

# add the predicted probabilities to 'alc'
alc <- mutate(alc, probability = probabilities)

# create logical vector indicating predictions
alc <- mutate(alc, prediction = probabilities > 0.5)

# table the target variable versus the predictions
table(high_use = alc$high_use, prediction = alc$prediction) %>% prop.table() %>% addmargins()
##         prediction
## high_use      FALSE       TRUE        Sum
##    FALSE 0.65945946 0.04054054 0.70000000
##    TRUE  0.24594595 0.05405405 0.30000000
##    Sum   0.90540541 0.09459459 1.00000000
# plot 'high_use' versus 'probability'
ggplot(alc, aes(x = high_use, y = probability)) + geom_point()

# define a loss function (mean prediction error)
loss_func <- function(class, prob) {
  n_wrong <- abs(class - prob) > 0.5
  mean(n_wrong)
}

# call loss_func to compute the average number of wrong predictions in the (training) data
loss_func(class = alc$high_use, prob = alc$probability)
## [1] 0.2864865

The training error of the model is 0.286 = 28.6% of classifications are wrong. I guess this is not ideal, but better than gessing (50% chance of classification being right or wrong).

Exercise 7. BONUS

Bonus: Perform 10-fold cross-validation on your model. Does your model have better test set performance (smaller prediction error using 10-fold cross-validation) compared to the model introduced in the Exercise Set (which had about 0.26 error). Could you find such a model? 

# K-fold cross-validation
cv <- cv.glm(data = alc, cost = loss_func, glmfit = alc.glm, K = nrow(alc))

# average number of wrong predictions in the cross validation
cv$delta[1]
## [1] 0.2891892

The test set error (0.289) is not better than the training error (0.286).

Exercise 8 BONUS

Chapter 4

Exercise 1 - Load Boston data

data("Boston")

Ecercise 2 - Explore the structure and the dimensions of the Boston data set

str(Boston)
## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...
dim(Boston)
## [1] 506  14

The Boston data set consists of housing Values in the suburbs of Boston. There are in total 506 observations and 14 variables related to housing properties, such as if the house bounds Charles river (chas), average number of rooms (rm) and median value of owner-occupied homes (medv). Variables are either numerical or integer variables.

Ecercise 3 - Graphical overview and summaries of the variables in the data

# summarize variables
summary(Boston)
##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08205   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00
# calculate the correlation matrix and round it to see relationships between the variables
cor_matrix <- cor(Boston) %>% round(digits=2)

# visualize the correlation matrix
corrplot(cor_matrix, method="circle", type="upper", cl.pos = "b", tl.pos = "d", tl.cex = 0.6)

From the summary we can see for example that:
- The per capita crime rate by town ranges from 0.006 to 88.976, mean is 3.677.
- Most houses don’t bound Charles river (mean of the dummy variable is 0.069).
- Average number of rooms is 6.3 (ranges from 3.6 to 8.8).
- Age ranges from 2.9 to 100, mean is 68.6.
- There are on average 18.46 pupils per teacher (range 12.6 to 22).
- Median value of homes is 22.53 * 1000 = 22530 $ (range 5000 - 50000).

From the correlation plot we can see for example that:
- Age and dis (weighted mean of distances to five Boston employment centres) have strong negative correlation, so younger people live closer to center
- nox (nitrogen oxides concentration) and dis have strong negative correlation, so more NO close to the center
- nox and age have strong positive correlation, which makes sense if older people live further away from the centers that have bigger NO concentrations
- rad (accessibility to radial highways) and tax (full-value property-tax rate) have strong positive correlation, so houses with access to highway have higher full-value property-tax rate
- rm and medv have strong positive correlation, so houses with more rooms are more expensive
- crim and rad and tax have positive correlation, so more crimes in areas with access to highways and higher property-tax rate

Exercise 4 - Standardization of the data

Standardize the dataset and print out summaries of the scaled data. How did the variables change? 

# center and standardize variables
boston_scaled <- Boston %>% scale()

# summary of the scaled variables
summary(boston_scaled)
##       crim                 zn               indus              chas        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563   Min.   :-0.2723  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668   1st Qu.:-0.2723  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109   Median :-0.2723  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150   3rd Qu.:-0.2723  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202   Max.   : 3.6648  
##       nox                rm               age               dis         
##  Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331   Min.   :-1.2658  
##  1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366   1st Qu.:-0.8049  
##  Median :-0.1441   Median :-0.1084   Median : 0.3171   Median :-0.2790  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059   3rd Qu.: 0.6617  
##  Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164   Max.   : 3.9566  
##       rad               tax             ptratio            black        
##  Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047   Min.   :-3.9033  
##  1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876   1st Qu.: 0.2049  
##  Median :-0.5225   Median :-0.4642   Median : 0.2746   Median : 0.3808  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058   3rd Qu.: 0.4332  
##  Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372   Max.   : 0.4406  
##      lstat              medv        
##  Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 3.5453   Max.   : 2.9865
# change the object to data frame
boston_scaled <- boston_scaled %>% as.data.frame()

The ranges of the variables are clearly much more even now, and for example tax that ranged from 187 to 711 or black (1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town) that ranged from 0.32 to 396.9 won’t dominate in the LDA analysis.

Create a categorical variable of the crime rate in the Boston dataset (from the scaled crime rate). Use the quantiles as the break points in the categorical variable. Drop the old crime rate variable from the dataset.

# convert scaled crim into numerical
#boston_scaled$crim <- as.numeric(boston_scaled$crim)
# create a quantile vector of crim 
bins <- quantile(boston_scaled$crim)

# create a categorical variable 'crime'
values =c("low", "med_low", "med_high", "high")
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, label= values)

# look at the table of the new factor crime
crime %>% table()
## .
##      low  med_low med_high     high 
##      127      126      126      127
# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)

# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)

Divide the dataset to train and test sets, so that 80% of the data belongs to the train set.

# number of rows in the Boston dataset 
n <- nrow(boston_scaled)

# choose randomly 80% of the rows
ind <- sample(n,  size = n * 0.8)

# create train set
train <- boston_scaled[ind,]

# create test set 
test <- boston_scaled[-ind,]

Exercise 5 - Fitting LDA

Fit the linear discriminant analysis on the train set. Use the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables. Draw the LDA (bi)plot.

# fit linear discriminant analysis with crime as the target and other variables as predictors
lda.fit <- lda(crime ~., data = train)

# print the lda.fit object
lda.fit
## Call:
## lda(crime ~ ., data = train)
## 
## Prior probabilities of groups:
##       low   med_low  med_high      high 
## 0.2599010 0.2326733 0.2673267 0.2400990 
## 
## Group means:
##                   zn      indus        chas        nox         rm        age
## low       0.97915788 -0.8932648 -0.08484810 -0.8788398  0.4621143 -0.8839619
## med_low  -0.09473193 -0.2788102 -0.10479289 -0.5184700 -0.1213189 -0.3088975
## med_high -0.38242941  0.1596639  0.23803578  0.3238515  0.1045362  0.3701257
## high     -0.48724019  1.0149946  0.01179157  1.0495941 -0.4464242  0.7775432
##                 dis        rad        tax      ptratio       black       lstat
## low       0.8882755 -0.6876449 -0.7847304 -0.415851342  0.38432420 -0.79161452
## med_low   0.2696760 -0.5505852 -0.4778474 -0.005012942  0.31522724 -0.17340223
## med_high -0.3542845 -0.3959403 -0.3014839 -0.209985732  0.06743763  0.01156479
## high     -0.8470049  1.6596029  1.5294129  0.805778429 -0.75980957  0.82177979
##                 medv
## low       0.55571875
## med_low   0.01355214
## med_high  0.15248048
## high     -0.62904405
## 
## Coefficients of linear discriminants:
##                  LD1          LD2          LD3
## zn       0.098390254  0.791171849 -0.815768715
## indus   -0.009110923 -0.106335378  0.080907308
## chas    -0.063420520 -0.073438142 -0.005532377
## nox      0.303751858 -0.743829826 -1.459602362
## rm      -0.104165988 -0.118095513 -0.231952795
## age      0.306304077 -0.231561216 -0.167714347
## dis     -0.071643294 -0.240246419 -0.137014805
## rad      3.145015304  1.180478450 -0.252582095
## tax      0.045743263 -0.346297377  0.835349347
## ptratio  0.118137848  0.052480881 -0.209820253
## black   -0.165430275  0.008263127  0.121359688
## lstat    0.172622069 -0.311113644  0.195308907
## medv     0.182868914 -0.384626907 -0.276903959
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9497 0.0399 0.0104
# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  graphics::arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

# target classes as numeric
classes <- as.numeric(train$crime)

# plot the lda results 
plot(lda.fit, dimen = 2)
lda.arrows(lda.fit, myscale = 1)

Exercise 6 - Prediction

Save the crime categories from the test set and then remove the categorical crime variable from the test dataset. Then predict the classes with the LDA model on the test data. Cross tabulate the results with the crime categories from the test set. Comment on the results.

# save the correct classes from test data
correct_classes <- test$crime

# remove the crime variable from test data
test <- dplyr::select(test, -crime)

# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)

# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)
##           predicted
## correct    low med_low med_high high
##   low       12       9        1    0
##   med_low    6      21        5    0
##   med_high   1       2       15    0
##   high       0       0        1   29

In this case, LDA is pretty good at predicting the high crime rates:
- all 23/23 “high” correctly assigned as high
Prediction accuracy of other classes is not as good:
- 16/23 of “low” are correctly assigned as low, 6/23 as med_low, 1/23 as med_high and zero as high
- 19/29 of “med_low” are correctly assigned as med_low, 6/29 as med_high and 4/29 as low
- 19/27 of “med_high” are correctly assigned as med_high, 6/29 as med_low and 2 as high

Exercise 7 - K-means clustering

Reload the Boston dataset and standardize the dataset. Calculate the distances between the observations.

data("Boston")

# scale and convert into data.frame
boston_scaled <- as.data.frame(scale(Boston))

# euclidean distance matrix
dist_eu <- dist(boston_scaled)

# look at the summary of the distances
summary(dist_eu)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.1343  3.4625  4.8241  4.9111  6.1863 14.3970

Run k-means algorithm on the dataset. Investigate what is the optimal number of clusters and run the algorithm again. Visualize the clusters (for example with the pairs() or ggpairs() functions, where the clusters are separated with colors) and interpret the results

# k-means clustering with clusters 1 to 5 and plotting
for (i in 1:5) {
  km <- kmeans(Boston, centers = i)
  pairs(Boston[, c("rm", "age", "dis", "crim")], col = km$cluster)
}

K = 4 an K = 5 seem to be too many, because the clusters overlap. 2 or 3 is maybe more suitable.

Chapter 5

Exercise 1 - Graphical overview

Move the country names to rownames. Show a graphical overview of the data and show summaries of the variables in the data. Describe and interpret the outputs, commenting on the distributions of the variables and the relationships between them.

human <- read.csv('./data/human.csv', header = T)
dim(human)
## [1] 155   9
human <- column_to_rownames(human, "Country")
head(human)
##             Sec_edu_ratio LFPR_ratio Edu_years Life_exp   GNI Mat_mort
## Norway          1.0072389  0.8908297      17.5     81.6 64992        4
## Australia       0.9968288  0.8189415      20.2     82.4 42261        6
## Switzerland     0.9834369  0.8251001      15.8     83.0 56431        6
## Denmark         0.9886128  0.8840361      18.7     80.2 44025        5
## Netherlands     0.9690608  0.8286119      17.9     81.6 45435        6
## Germany         0.9927835  0.8072289      16.5     80.9 43919        7
##             Adol_birth_rate Rep_parl
## Norway                  7.8     39.6
## Australia              12.1     30.5
## Switzerland             1.9     28.5
## Denmark                 5.1     38.0
## Netherlands             6.2     36.9
## Germany                 3.8     36.9
dim(human)
## [1] 155   8
ggpairs(human, progress = FALSE)

cor_matrix <- cor(human) %>% round(digits=2)
corrplot(cor_matrix, method="circle", type="upper", cl.pos = "b", tl.pos = "d", tl.cex = 0.6)

From the ggpairs plot we can see that many variables are not normally distributed, such as GNI and Mat_mort. Many of the variables are strongly correlated. From the correlation plot we can confirm that many variables have strong positive and negative correlations, but that the variables LFPR_ratio and Rep_parl don’t seem to correlate as much with the other variables.

Excercise 2 - PCA on non-standardized data

Perform principal component analysis (PCA) on the raw (non-standardized) human data. Show the variability captured by the principal components. Draw a biplot displaying the observations by the first two principal components (PC1 coordinate in x-axis, PC2 coordinate in y-axis), along with arrows representing the original variables

# perform PCA on non-standardized data
pca_human <- prcomp(human)
summary(pca_human) # proportion of variance and cumulative proportion captured by the components 
## Importance of components:
##                              PC1      PC2   PC3   PC4   PC5   PC6    PC7    PC8
## Standard deviation     1.854e+04 185.5219 25.19 11.45 3.766 1.566 0.1912 0.1591
## Proportion of Variance 9.999e-01   0.0001  0.00  0.00 0.000 0.000 0.0000 0.0000
## Cumulative Proportion  9.999e-01   1.0000  1.00  1.00 1.000 1.000 1.0000 1.0000
# draw biplot
biplot(pca_human, choices = 1:2)

Since GNI has a lot wider range of values than the other variables and the data hasn’t been standardized, GNI dominates the analysis. Pretty much all the variability in the data is explained by GNI alone and the first principal component explains the majority of the variation.

Excercise 3 - PCA on standardized data

Standardize the variables in the human data and repeat the above analysis. Interpret the results of both analysis (with and without standardizing). Are the results different? Why or why not? Include captions (brief descriptions) in your plots where you describe the results by using not just your variable names, but the actual phenomena they relate to.

# standardize data
human_std <- scale(human)

# perform PCA on standardized data
pca_human_std <- prcomp(human_std)
summary(pca_human_std)
## Importance of components:
##                           PC1    PC2     PC3     PC4     PC5     PC6     PC7
## Standard deviation     2.0708 1.1397 0.87505 0.77886 0.66196 0.53631 0.45900
## Proportion of Variance 0.5361 0.1624 0.09571 0.07583 0.05477 0.03595 0.02634
## Cumulative Proportion  0.5361 0.6984 0.79413 0.86996 0.92473 0.96069 0.98702
##                            PC8
## Standard deviation     0.32224
## Proportion of Variance 0.01298
## Cumulative Proportion  1.00000
# create and print out a summary of pca_human
s <- summary(pca_human_std)

# rounded percentanges of variance captured by each PC
pca_pr <- round(1*s$importance[2, ], digits = 5)

# create object pc_lab to be used as axis labels
pc_lab <- paste0(names(pca_pr), " (", pca_pr * 100, "%)")

# draw a biplot Use the first value of the `pc_lab` vector as the label for the x-axis and the second value as the label for the y-axis

biplot(pca_human_std, cex = c(0.8, 1), col = c("grey40", "deeppink2"), xlab = pc_lab[1], ylab = pc_lab[2])

The biplot looks different, since after scaling the data, the variables contribute equally to the analysis. The biplot is much more even now. The first principal component now explains 54% of the variation.

Excercise 4 - Interpretations of the first two principal component dimensions

Give your personal interpretations of the first two principal component dimensions based on the biplot drawn after PCA on the standardized human data.

Biplot shows how strongly each characteristic influences a principal component. We can see that PC1 is is influenced by factors such as years of education and life expectancy, which are positively correlated with each other, as well as Mat_mort, which is negatively correlated with the other 2. PC2 is influenced by LFPR_ratio and Rep_parl, which are also positively correlated with each other.

Excercise 5 - Visualization of Tea data

Load the tea dataset and convert its character variables to factors. Explore the data briefly: look at the structure and the dimensions of the data. Use View(tea) to browse its contents, and visualize the data.

tea <- read.csv("https://raw.githubusercontent.com/KimmoVehkalahti/Helsinki-Open-Data-Science/master/datasets/tea.csv", stringsAsFactors = TRUE)
str(tea)
## 'data.frame':    300 obs. of  36 variables:
##  $ breakfast       : Factor w/ 2 levels "breakfast","Not.breakfast": 1 1 2 2 1 2 1 2 1 1 ...
##  $ tea.time        : Factor w/ 2 levels "Not.tea time",..: 1 1 2 1 1 1 2 2 2 1 ...
##  $ evening         : Factor w/ 2 levels "evening","Not.evening": 2 2 1 2 1 2 2 1 2 1 ...
##  $ lunch           : Factor w/ 2 levels "lunch","Not.lunch": 2 2 2 2 2 2 2 2 2 2 ...
##  $ dinner          : Factor w/ 2 levels "dinner","Not.dinner": 2 2 1 1 2 1 2 2 2 2 ...
##  $ always          : Factor w/ 2 levels "always","Not.always": 2 2 2 2 1 2 2 2 2 2 ...
##  $ home            : Factor w/ 2 levels "home","Not.home": 1 1 1 1 1 1 1 1 1 1 ...
##  $ work            : Factor w/ 2 levels "Not.work","work": 1 1 2 1 1 1 1 1 1 1 ...
##  $ tearoom         : Factor w/ 2 levels "Not.tearoom",..: 1 1 1 1 1 1 1 1 1 2 ...
##  $ friends         : Factor w/ 2 levels "friends","Not.friends": 2 2 1 2 2 2 1 2 2 2 ...
##  $ resto           : Factor w/ 2 levels "Not.resto","resto": 1 1 2 1 1 1 1 1 1 1 ...
##  $ pub             : Factor w/ 2 levels "Not.pub","pub": 1 1 1 1 1 1 1 1 1 1 ...
##  $ Tea             : Factor w/ 3 levels "black","Earl Grey",..: 1 1 2 2 2 2 2 1 2 1 ...
##  $ How             : Factor w/ 4 levels "alone","lemon",..: 1 3 1 1 1 1 1 3 3 1 ...
##  $ sugar           : Factor w/ 2 levels "No.sugar","sugar": 2 1 1 2 1 1 1 1 1 1 ...
##  $ how             : Factor w/ 3 levels "tea bag","tea bag+unpackaged",..: 1 1 1 1 1 1 1 1 2 2 ...
##  $ where           : Factor w/ 3 levels "chain store",..: 1 1 1 1 1 1 1 1 2 2 ...
##  $ price           : Factor w/ 6 levels "p_branded","p_cheap",..: 4 6 6 6 6 3 6 6 5 5 ...
##  $ age             : int  39 45 47 23 48 21 37 36 40 37 ...
##  $ sex             : Factor w/ 2 levels "F","M": 2 1 1 2 2 2 2 1 2 2 ...
##  $ SPC             : Factor w/ 7 levels "employee","middle",..: 2 2 4 6 1 6 5 2 5 5 ...
##  $ Sport           : Factor w/ 2 levels "Not.sportsman",..: 2 2 2 1 2 2 2 2 2 1 ...
##  $ age_Q           : Factor w/ 5 levels "+60","15-24",..: 4 5 5 2 5 2 4 4 4 4 ...
##  $ frequency       : Factor w/ 4 levels "+2/day","1 to 2/week",..: 3 3 1 3 1 3 4 2 1 1 ...
##  $ escape.exoticism: Factor w/ 2 levels "escape-exoticism",..: 2 1 2 1 1 2 2 2 2 2 ...
##  $ spirituality    : Factor w/ 2 levels "Not.spirituality",..: 1 1 1 2 2 1 1 1 1 1 ...
##  $ healthy         : Factor w/ 2 levels "healthy","Not.healthy": 1 1 1 1 2 1 1 1 2 1 ...
##  $ diuretic        : Factor w/ 2 levels "diuretic","Not.diuretic": 2 1 1 2 1 2 2 2 2 1 ...
##  $ friendliness    : Factor w/ 2 levels "friendliness",..: 2 2 1 2 1 2 2 1 2 1 ...
##  $ iron.absorption : Factor w/ 2 levels "iron absorption",..: 2 2 2 2 2 2 2 2 2 2 ...
##  $ feminine        : Factor w/ 2 levels "feminine","Not.feminine": 2 2 2 2 2 2 2 1 2 2 ...
##  $ sophisticated   : Factor w/ 2 levels "Not.sophisticated",..: 1 1 1 2 1 1 1 2 2 1 ...
##  $ slimming        : Factor w/ 2 levels "No.slimming",..: 1 1 1 1 1 1 1 1 1 1 ...
##  $ exciting        : Factor w/ 2 levels "exciting","No.exciting": 2 1 2 2 2 2 2 2 2 2 ...
##  $ relaxing        : Factor w/ 2 levels "No.relaxing",..: 1 1 2 2 2 2 2 2 2 2 ...
##  $ effect.on.health: Factor w/ 2 levels "effect on health",..: 2 2 2 2 2 2 2 2 2 2 ...
# There are no character variables, so I'm not sure what should be converted here..
summary(tea)
##          breakfast           tea.time          evening          lunch    
##  breakfast    :144   Not.tea time:131   evening    :103   lunch    : 44  
##  Not.breakfast:156   tea time    :169   Not.evening:197   Not.lunch:256  
##                                                                          
##                                                                          
##                                                                          
##                                                                          
##                                                                          
##         dinner           always          home           work    
##  dinner    : 21   always    :103   home    :291   Not.work:213  
##  Not.dinner:279   Not.always:197   Not.home:  9   work    : 87  
##                                                                 
##                                                                 
##                                                                 
##                                                                 
##                                                                 
##         tearoom           friends          resto          pub     
##  Not.tearoom:242   friends    :196   Not.resto:221   Not.pub:237  
##  tearoom    : 58   Not.friends:104   resto    : 79   pub    : 63  
##                                                                   
##                                                                   
##                                                                   
##                                                                   
##                                                                   
##         Tea         How           sugar                     how     
##  black    : 74   alone:195   No.sugar:155   tea bag           :170  
##  Earl Grey:193   lemon: 33   sugar   :145   tea bag+unpackaged: 94  
##  green    : 33   milk : 63                  unpackaged        : 36  
##                  other:  9                                          
##                                                                     
##                                                                     
##                                                                     
##                   where                 price          age        sex    
##  chain store         :192   p_branded      : 95   Min.   :15.00   F:178  
##  chain store+tea shop: 78   p_cheap        :  7   1st Qu.:23.00   M:122  
##  tea shop            : 30   p_private label: 21   Median :32.00          
##                             p_unknown      : 12   Mean   :37.05          
##                             p_upscale      : 53   3rd Qu.:48.00          
##                             p_variable     :112   Max.   :90.00          
##                                                                          
##            SPC               Sport       age_Q          frequency  
##  employee    :59   Not.sportsman:121   +60  :38   +2/day     :127  
##  middle      :40   sportsman    :179   15-24:92   1 to 2/week: 44  
##  non-worker  :64                       25-34:69   1/day      : 95  
##  other worker:20                       35-44:40   3 to 6/week: 34  
##  senior      :35                       45-59:61                    
##  student     :70                                                   
##  workman     :12                                                   
##              escape.exoticism           spirituality        healthy   
##  escape-exoticism    :142     Not.spirituality:206   healthy    :210  
##  Not.escape-exoticism:158     spirituality    : 94   Not.healthy: 90  
##                                                                       
##                                                                       
##                                                                       
##                                                                       
##                                                                       
##          diuretic             friendliness            iron.absorption
##  diuretic    :174   friendliness    :242   iron absorption    : 31   
##  Not.diuretic:126   Not.friendliness: 58   Not.iron absorption:269   
##                                                                      
##                                                                      
##                                                                      
##                                                                      
##                                                                      
##          feminine             sophisticated        slimming          exciting  
##  feminine    :129   Not.sophisticated: 85   No.slimming:255   exciting   :116  
##  Not.feminine:171   sophisticated    :215   slimming   : 45   No.exciting:184  
##                                                                                
##                                                                                
##                                                                                
##                                                                                
##                                                                                
##         relaxing              effect.on.health
##  No.relaxing:113   effect on health   : 66    
##  relaxing   :187   No.effect on health:234    
##                                               
##                                               
##                                               
##                                               
## 
# make a smaller data set of 'tea'
keep_columns <- c("breakfast", "evening", "work", "Tea", "sugar", "sex", "Sport", "relaxing", "effect.on.health")

# select the 'keep_columns' to create a new dataset
tea_time <- dplyr::select(tea, keep_columns)
## Warning: Using an external vector in selections was deprecated in tidyselect
## 1.1.0.

## Warning: Please use `all_of()` or `any_of()` instead.

## Warning: # Was:

## Warning: data %>% select(keep_columns)

## Warning:

## Warning: # Now:

## Warning: data %>% select(all_of(keep_columns))

## Warning:

## Warning: See <https://tidyselect.r-lib.org/reference/faq-external-vector.html>.
# ggplot
pivot_longer(tea_time, cols = everything()) %>% 
  ggplot(aes(value)) + facet_wrap("name", scales = "free") + geom_bar()

Exercise 5 - Multiple Correspondence Analysis (MCA)

Use Multiple Correspondence Analysis (MCA) on the tea data (or on just certain columns of the data, it is up to you!). Interpret the results of the MCA and draw at least the variable biplot of the analysis. You can also explore other plotting options for MCA. Comment on the output of the plots.

# multiple correspondence analysis
mca <- MCA(tea_time, graph = FALSE)

# summary of the model
summary(mca)
## 
## Call:
## MCA(X = tea_time, graph = FALSE) 
## 
## 
## Eigenvalues
##                        Dim.1   Dim.2   Dim.3   Dim.4   Dim.5   Dim.6   Dim.7
## Variance               0.162   0.145   0.134   0.132   0.119   0.102   0.091
## % of var.             14.570  13.059  12.047  11.859  10.720   9.215   8.149
## Cumulative % of var.  14.570  27.629  39.676  51.534  62.254  71.470  79.619
##                        Dim.8   Dim.9  Dim.10
## Variance               0.083   0.078   0.066
## % of var.              7.453   7.008   5.919
## Cumulative % of var.  87.072  94.081 100.000
## 
## Individuals (the 10 first)
##                  Dim.1    ctr   cos2    Dim.2    ctr   cos2    Dim.3    ctr
## 1             |  0.007  0.000  0.000 |  0.199  0.091  0.035 |  0.407  0.412
## 2             |  0.500  0.515  0.242 | -0.187  0.080  0.034 |  0.595  0.880
## 3             | -0.303  0.190  0.092 | -0.444  0.452  0.196 | -0.082  0.017
## 4             | -0.230  0.109  0.065 |  0.383  0.337  0.181 | -0.314  0.246
## 5             | -0.503  0.522  0.288 | -0.009  0.000  0.000 |  0.387  0.373
## 6             | -0.115  0.027  0.019 |  0.192  0.085  0.052 |  0.071  0.013
## 7             | -0.154  0.049  0.033 | -0.086  0.017  0.010 |  0.324  0.262
## 8             | -0.160  0.053  0.024 |  0.142  0.046  0.019 |  0.593  0.877
## 9             | -0.154  0.049  0.033 | -0.086  0.017  0.010 |  0.324  0.262
## 10            | -0.179  0.066  0.026 |  0.205  0.096  0.034 |  0.849  1.796
##                 cos2  
## 1              0.146 |
## 2              0.342 |
## 3              0.007 |
## 4              0.122 |
## 5              0.170 |
## 6              0.007 |
## 7              0.145 |
## 8              0.334 |
## 9              0.145 |
## 10             0.578 |
## 
## Categories (the 10 first)
##                   Dim.1     ctr    cos2  v.test     Dim.2     ctr    cos2
## breakfast     |  -0.074   0.182   0.005  -1.235 |  -0.495   9.012   0.226
## Not.breakfast |   0.069   0.168   0.005   1.235 |   0.457   8.319   0.226
## evening       |  -0.830  16.240   0.360 -10.380 |   0.173   0.783   0.016
## Not.evening   |   0.434   8.491   0.360  10.380 |  -0.090   0.409   0.016
## Not.work      |   0.019   0.018   0.001   0.522 |   0.442  10.625   0.478
## work          |  -0.047   0.044   0.001  -0.522 |  -1.082  26.013   0.478
## black         |   0.224   0.850   0.016   2.217 |   0.165   0.512   0.009
## Earl Grey     |  -0.227   2.280   0.093  -5.277 |  -0.318   4.982   0.182
## green         |   0.826   5.157   0.084   5.024 |   1.491  18.716   0.275
## No.sugar      |   0.550  10.742   0.324   9.840 |  -0.196   1.515   0.041
##                v.test     Dim.3     ctr    cos2  v.test  
## breakfast      -8.226 |   0.434   7.490   0.174   7.203 |
## Not.breakfast   8.226 |  -0.400   6.914   0.174  -7.203 |
## evening         2.158 |   0.136   0.528   0.010   1.701 |
## Not.evening    -2.158 |  -0.071   0.276   0.010  -1.701 |
## Not.work       11.961 |   0.110   0.717   0.030   2.983 |
## work          -11.961 |  -0.270   1.754   0.030  -2.983 |
## black           1.629 |   1.349  37.243   0.596  13.345 |
## Earl Grey      -7.385 |  -0.494  13.026   0.440 -11.469 |
## green           9.062 |  -0.136   0.169   0.002  -0.826 |
## No.sugar       -3.498 |   0.458   8.987   0.224   8.184 |
## 
## Categorical variables (eta2)
##                    Dim.1 Dim.2 Dim.3  
## breakfast        | 0.005 0.226 0.174 |
## evening          | 0.360 0.016 0.010 |
## work             | 0.001 0.478 0.030 |
## Tea              | 0.121 0.316 0.608 |
## sugar            | 0.324 0.041 0.224 |
## sex              | 0.101 0.204 0.026 |
## Sport            | 0.126 0.015 0.025 |
## relaxing         | 0.379 0.002 0.091 |
## effect.on.health | 0.041 0.008 0.018 |
# visualize the model
plot(mca, invisible=c("ind"), graph.type = "classic", habillage = "quali")

plot(mca, invisible=c("var"), graph.type = "classic", habillage = "quali")

The distance between variables gives a measure of their similarity (or dissimilarity). For example Relaxing and evening seem to contribute most to Dim 1, whereas variables work and Tea contribute greatly to Dim2.

From the individuals’ scatterplot we can see that there is no particular group of individuals, the scatterplot is quite homogeneous.

Chapter 6

1. Meet and Repeat: PART I using the RATS data

RATS dataset contains data from a nutrition study conducted in three groups of rats (Crowder and Hand, 1990). The three groups were put on different diets, and each animal’s body weight (grams) was recorded repeatedly (approximately weekly, except in week seven when two recordings were taken) over a 9-week period. The question of most interest is whether the growth profiles of the three groups differ.

RATS data has been converted into long format in meet_and_repeat.R, so I import the data:

RATSL <- read.table("./data/RATSL.txt", header = T)

# convert ID and Group into factors
RATSL$ID <- factor(RATSL$ID)
RATSL$Group <- factor(RATSL$Group)

# Draw ggplot with `Time` on the x-axis and `Weight` on the y-axis
ggplot(RATSL, aes(x = Time, y = Weight, linetype = ID)) +
  geom_line() +
  scale_linetype_manual(values = rep(1:10, times=20)) +
  facet_grid(. ~ Group, labeller = label_both) +
  theme(legend.position = "none") + 
  scale_y_continuous(limits = c(min(RATSL$Weight), max(RATSL$Weight)))

From the plot we can see that the starting weight differs a lot for the three groups of rats. In all groups, weight increases during the study time, more so in groups 2 and 3. There is also one rat in group 2 that differs from the other rats in the group.

Next, we inspect the tracking to see if rats that have higher weight values at the beginning tend to have higher values throughout the study.

The tracking phenomenon can be seen more clearly in a plot of the standardized values of each observation, i.e., the values obtained by subtracting the relevant occasion mean from the original observation and then dividing by the corresponding visit standard deviation:

\[standardised(x) = \frac{x - mean(x)}{ sd(x)}\]

# Group by WD and standardise the variable Weight
RATSL <- RATSL %>%
  group_by(WD) %>%
  mutate(stdweight = scale(Weight)) %>%
  ungroup()

# Plot again with the standardised Weight
ggplot(RATSL, aes(x = Time, y = stdweight, linetype = ID)) +
  geom_line() +
  scale_linetype_manual(values = rep(1:10, times=20)) +
  facet_grid(. ~ Group, labeller = label_both) +
  scale_y_continuous(name = "standardized weight")

Let’sproduce graphs showing average (mean) profiles for each group along with some indication of the variation of the observations at each time point, in this case the standard error of mean

\[se = \frac{sd(x)}{\sqrt{n}}\]

# Summary data with mean and standard error of Weight by Group and Time 
RATSS <- RATSL %>%
  group_by(Group, Time) %>%
  summarise( mean = mean(Weight), se = sd(Weight) ) %>%
  ungroup()
## `summarise()` has grouped output by 'Group'. You can override using the
## `.groups` argument.
# Plot the mean profiles
ggplot(RATSS, aes(x = Time, y = mean, linetype = Group, shape = Group)) +
  geom_line() +
  scale_linetype_manual(values = c(1,2,3)) +
  geom_point(size=3) +
  scale_shape_manual(values = c(1,2,3)) +
  geom_errorbar(aes(ymin=mean-se, ymax=mean+se, linetype="1"), width=0.3) +
  theme(legend.position = c(0.9,0.4)) +
  scale_y_continuous(name = "mean(Weight) +/- se(Weight)")

We can see that the measurements are not independent. The mean of group 1 differs from that of groups 2 and 3, whereas the standard errors of groups 2 and 3 overlap. Variation of weight in rat groups 2 and 3 is bigger than in group 1 (there seemed to be outliers in group 2 and 3 in the plot in previous step increasing variation). Increase in weight seems to be only slightly bigger in group 2.

Next let’s check for these outliers.

# Create a summary data by Group and ID with mean as the summary variable
RATSL8S <- RATSL %>%
  group_by(Group, ID) %>%
  summarise(mean=mean(Weight) ) %>%
  ungroup()
## `summarise()` has grouped output by 'Group'. You can override using the
## `.groups` argument.
ggplot(RATSL8S, aes(x = Group, y = mean)) +
  geom_boxplot() +
  stat_summary(fun = "mean", geom = "point", shape=23, size=4, fill = "white") +
  scale_y_continuous(name = "mean(Weight), Days 1-64") +
  ggtitle("Mean weight in the groups, all rats included") +
  theme(plot.title = element_text(hjust = 0.5))

There seems to be a clear outlier in group 2. Maybe the rats with lower mean weights in groups 1 and 3 could also be considered outliers, but here I only consider the rat in group 2 as an outlier and remove it from the data.

# Create a new data by filtering the outlier and adjust the ggplot code the draw the plot again with the new data
RATSL8S1 <- RATSL8S[-12, ]

ggplot(RATSL8S1, aes(x = Group, y = mean)) +
  geom_boxplot() +
  stat_summary(fun = "mean", geom = "point", shape=23, size=4, fill = "white") +
  scale_y_continuous(name = "mean(Weight), Days 1-64") +
  ggtitle("Mean weight in the groups, outlier in group 2 removed") +
  theme(plot.title = element_text(hjust = 0.5))

Graphical presentations indicated a lack of difference in the three diets, but let’s perform a t-test to check for difference.

Baseline measurements of the outcome variable in a longitudinal study are often correlated with the chosen summary measure and using such measures in the analysis can often lead to substantial gains in precision when used appropriately as a covariate in an analysis of covariance. We can illustrate the analysis on the data using the weight corresponding to time zero taken prior to the start of study as the baseline covariate.

# Add the baseline from the original data as a new variable to the summary data
RATS <- read.table("https://raw.githubusercontent.com/KimmoVehkalahti/MABS/master/Examples/data/rats.txt", header = TRUE, sep = '\t')

RATSL8S2 <- RATSL8S %>%
 mutate(baseline = RATS$WD1)

# Fit the linear model with the mean as the response 
fit <- lm(mean ~ baseline + Group, data = RATSL8S2)
summary(fit)
## 
## Call:
## lm(formula = mean ~ baseline + Group, data = RATSL8S2)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -21.732  -3.812   1.991   6.889  13.455 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 30.14886   19.88779   1.516   0.1554    
## baseline     0.93194    0.07793  11.959 5.02e-08 ***
## Group2      31.68866   17.11189   1.852   0.0888 .  
## Group3      21.52296   21.13931   1.018   0.3287    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 10.62 on 12 degrees of freedom
## Multiple R-squared:  0.9947, Adjusted R-squared:  0.9933 
## F-statistic: 747.8 on 3 and 12 DF,  p-value: 6.636e-14
confint(fit)
##                   2.5 %    97.5 %
## (Intercept) -13.1829034 73.480627
## baseline      0.7621505  1.101726
## Group2       -5.5949497 68.972271
## Group3      -24.5356460 67.581567
# Compute the analysis of variance table for the fitted model with anova()
# anova(fit)
anova(fit)
## Analysis of Variance Table
## 
## Response: mean
##           Df Sum Sq Mean Sq   F value    Pr(>F)    
## baseline   1 252125  252125 2237.0655 5.217e-15 ***
## Group      2    726     363    3.2219   0.07586 .  
## Residuals 12   1352     113                        
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Summary for the linear model shows that the baseline of the groups is associated with weight and that the association is statistically significant (p-value 5.02e-08), whereas the group (=diet) is not associated with weight, although there is slight indication of association for group 2 (p-value 0.0888).The 95% confidence intervals are also wide and include the zero, so there is no evidence that the group differences differ from zero.

Anova confirms the lack of evidence for a group difference (p-value 0.07586).

2. Meet and Repeat: PART II using the BPRS data

BPRS data consists of 40 male subjects randomly assigned to one of two treatment groups and each subject was rated on the brief psychiatric rating scale (BPRS) measured before treatment began (week 0) and then at weekly intervals for eight weeks. The BPRS assesses the level of 18 symptom constructs such as hostility, suspiciousness, hallucinations and grandiosity; each of these is rated from one (not present) to seven (extremely severe). The scale is used to evaluate patients suspected of having schizophrenia.

BPRS data has been converted into long format in meet_and_repeat.R, so I import the data:

BPRSL <- read.table("./data/BPRSL.txt", header = T)

# convert ID and Group into factors
BPRSL$treatment <- factor(BPRSL$treatment)
BPRSL$subject <- factor(BPRSL$subject)


# Draw the plot
ggplot(BPRSL, aes(x = week, y = bprs, linetype = subject)) +
  geom_line() +
  scale_linetype_manual(values = rep(1:10, times=4)) +
  facet_grid(. ~ treatment, labeller = label_both) +
  theme(legend.position = "none") + 
  scale_y_continuous(limits = c(min(BPRSL$bprs), max(BPRSL$bprs)))

The bprs values seem to decrease in both treatment groups during the treatment weeks 1-8, but there doesn’t seem to be significant differences between the groups.

Let’s fit a multiple linear regression model to see if this is indeed correct.

# fit a multiple linear regression model with `pbrs` as response and `week` and `treatment` as explanatory variables

# create a regression model RATS_reg
BPRSL_reg <- lm(data=BPRSL, bprs ~ week + treatment)

# print out a summary of the model
summary(BPRSL_reg)
## 
## Call:
## lm(formula = bprs ~ week + treatment, data = BPRSL)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -22.454  -8.965  -3.196   7.002  50.244 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  46.4539     1.3670  33.982   <2e-16 ***
## week         -2.2704     0.2524  -8.995   <2e-16 ***
## treatment2    0.5722     1.3034   0.439    0.661    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 12.37 on 357 degrees of freedom
## Multiple R-squared:  0.1851, Adjusted R-squared:  0.1806 
## F-statistic: 40.55 on 2 and 357 DF,  p-value: < 2.2e-16

In the summary we can see that time (week) is statistically significantly (p-value <2e-16) negatively correlated with pbrs value, so during treatment weeks 1-8 the bprs value goes down. However, the type of treatment (treatment) is not statistically significant, so the type of treatment doesn’t have an effect on the decrease of pbrs value, as could be inspected from the plot above.

The lm model assumes independence of the repeated measures of bprs, and this assumption is highly unlikely. So, now we will move on to consider both some more appropriate graphics and appropriate models.

To begin the more formal analysis of the bprs data, we will first fit the random intercept model for the same two explanatory variables: week and treatment. Fitting a random intercept model allows the linear regression fit for each study subject to differ in intercept from other subjects. So what we do is we combine the variance components and the single level regression model and we get a random intercept model.

# Create a random intercept model as subject as the random effect
BPRS_ref <- lmer(bprs ~ week + treatment + (1 | subject), data = BPRSL, REML = FALSE)

# Print the summary of the model
summary(BPRS_ref)
## Linear mixed model fit by maximum likelihood  ['lmerMod']
## Formula: bprs ~ week + treatment + (1 | subject)
##    Data: BPRSL
## 
##      AIC      BIC   logLik deviance df.resid 
##   2748.7   2768.1  -1369.4   2738.7      355 
## 
## Scaled residuals: 
##     Min      1Q  Median      3Q     Max 
## -3.0481 -0.6749 -0.1361  0.4813  3.4855 
## 
## Random effects:
##  Groups   Name        Variance Std.Dev.
##  subject  (Intercept)  47.41    6.885  
##  Residual             104.21   10.208  
## Number of obs: 360, groups:  subject, 20
## 
## Fixed effects:
##             Estimate Std. Error t value
## (Intercept)  46.4539     1.9090  24.334
## week         -2.2704     0.2084 -10.896
## treatment2    0.5722     1.0761   0.532
## 
## Correlation of Fixed Effects:
##            (Intr) week  
## week       -0.437       
## treatment2 -0.282  0.000

We see here again that treatment and time don’t correlate with each other. The t value for treatment is close to zero (0.532), so it’s p-value is large, whereas the t-statistic for week is far from zero (-10.896) with corresponding small p-value, which indicates that it is unlikely to get such results by chance and that the null hypothesis of no correlation between time and bprs should not be rejected.

Now we can move on to fit the random intercept and random slope model to the BPRS data. Fitting a random intercept and random slope model allows the linear regression fits for each individual to differ in intercept but also in slope. This way it is possible to account for the individual differences in the bprs profiles, but also the effect of time.

# create a random intercept and random slope model with `week` and `subject` as the random effects
BPRS_ref1 <- lmer(bprs ~ week + treatment + (week | subject), data = BPRSL, REML = FALSE)

# print a summary of the model
summary(BPRS_ref1)
## Linear mixed model fit by maximum likelihood  ['lmerMod']
## Formula: bprs ~ week + treatment + (week | subject)
##    Data: BPRSL
## 
##      AIC      BIC   logLik deviance df.resid 
##   2745.4   2772.6  -1365.7   2731.4      353 
## 
## Scaled residuals: 
##     Min      1Q  Median      3Q     Max 
## -2.8919 -0.6194 -0.0691  0.5531  3.7976 
## 
## Random effects:
##  Groups   Name        Variance Std.Dev. Corr 
##  subject  (Intercept) 64.8222  8.0512        
##           week         0.9609  0.9802   -0.51
##  Residual             97.4305  9.8707        
## Number of obs: 360, groups:  subject, 20
## 
## Fixed effects:
##             Estimate Std. Error t value
## (Intercept)  46.4539     2.1052  22.066
## week         -2.2704     0.2977  -7.626
## treatment2    0.5722     1.0405   0.550
## 
## Correlation of Fixed Effects:
##            (Intr) week  
## week       -0.582       
## treatment2 -0.247  0.000
# Compute the ANOVA analysis of variance tables of the models 
anova(BPRS_ref1, BPRS_ref)
## Data: BPRSL
## Models:
## BPRS_ref: bprs ~ week + treatment + (1 | subject)
## BPRS_ref1: bprs ~ week + treatment + (week | subject)
##           npar    AIC    BIC  logLik deviance  Chisq Df Pr(>Chisq)  
## BPRS_ref     5 2748.7 2768.1 -1369.4   2738.7                       
## BPRS_ref1    7 2745.4 2772.6 -1365.7   2731.4 7.2721  2    0.02636 *
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

BPRS_ref1 (a random intercept and random slope model) is a better fit than BPRS_ref (a random intercept model) in terms of chi-squared statistics (7.2721) and p-value of the likelihood ratio test (0.02636).

Finally, we can fit a random intercept and slope model that allows for a treatment group × time interaction, even though we saw that treatment and time don’t correlate with each other.

# Write the same model as in the previous exercise but add `week` * `treatment` interaction.
BPRS_ref2 <- lmer(bprs ~ week + treatment + (week | subject) + (week * treatment), data = BPRSL, REML = FALSE)

# print a summary of the model
summary(BPRS_ref2)
## Linear mixed model fit by maximum likelihood  ['lmerMod']
## Formula: bprs ~ week + treatment + (week | subject) + (week * treatment)
##    Data: BPRSL
## 
##      AIC      BIC   logLik deviance df.resid 
##   2744.3   2775.4  -1364.1   2728.3      352 
## 
## Scaled residuals: 
##     Min      1Q  Median      3Q     Max 
## -3.0512 -0.6271 -0.0768  0.5288  3.9260 
## 
## Random effects:
##  Groups   Name        Variance Std.Dev. Corr 
##  subject  (Intercept) 64.9964  8.0620        
##           week         0.9687  0.9842   -0.51
##  Residual             96.4707  9.8220        
## Number of obs: 360, groups:  subject, 20
## 
## Fixed effects:
##                 Estimate Std. Error t value
## (Intercept)      47.8856     2.2521  21.262
## week             -2.6283     0.3589  -7.323
## treatment2       -2.2911     1.9090  -1.200
## week:treatment2   0.7158     0.4010   1.785
## 
## Correlation of Fixed Effects:
##             (Intr) week   trtmn2
## week        -0.650              
## treatment2  -0.424  0.469       
## wek:trtmnt2  0.356 -0.559 -0.840
# perform an ANOVA test on the two models
anova(BPRS_ref2, BPRS_ref1)
## Data: BPRSL
## Models:
## BPRS_ref1: bprs ~ week + treatment + (week | subject)
## BPRS_ref2: bprs ~ week + treatment + (week | subject) + (week * treatment)
##           npar    AIC    BIC  logLik deviance  Chisq Df Pr(>Chisq)  
## BPRS_ref1    7 2745.4 2772.6 -1365.7   2731.4                       
## BPRS_ref2    8 2744.3 2775.4 -1364.1   2728.3 3.1712  1    0.07495 .
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

As could be expected, the model with the week * treatment interaction variable is not statistically better than the model without interaction (chi-squared statistic 3.1712, p-value 0.07495). So here the addition of the interaction was not necessary and didn’t improve the model.